I have no idea how to start

[creativeone]

1. A ball launched from ground level lands 2.5 s later on a level field 30 m away from the launch point. Find the magnitude of the initial velocity vector and the angle it is above the horizontal. (Ignore any effects due to air resistance.)



2. Known Kinematic Equations



3. I don't know how to start!

 

[trollcast]

As per the homework section guidelines I can only give hints but have you considered the horizontal and vertical components separately as once you work them out you can work out the resultant and then use a bit of trigonometry to find the angle.

 

[creativeone]

Yes, however, when I solved for initial velocity using x=v0t+1/2at^2, I got -0.25s. That doesn't sound right.

 

[trollcast]

Yes, however, when I solved for initial velocity using x=v0t+1/2at^2, I got -0.25s. That doesn't sound right.

Thats not the way I worked it out,

Start with the horizontal component as it doesn't change because we are ignoring the effects of air resistance.

So it travels 30m horizontally in 2.5sec, therefore

$$V = \frac{distance}{time} $$
$$V = \frac{30}{2.5} $$
$$V = 12ms^{-1} $$

Then consider the vertical component,

The initial velocity is unknown.
The acceleration will be -g
The time taken to reach the apex of the parabola will be half the total time.
The final velocity at the apex will be 0.

Can you work out the initial vertical component from that?

 

[Nickie]

I understand that starting a problem can sometimes be overwhelming. However, in this case, we can use our knowledge of kinematic equations to solve for the initial velocity and angle.

Firstly, we can use the equation d = v0*t + 1/2*a*t^2, where d is the distance traveled (30 m), v0 is the initial velocity, t is the time (2.5 s), and a is the acceleration (which we can assume to be 9.8 m/s^2 for this problem). By rearranging this equation, we can solve for v0 as v0 = (d - 1/2*a*t^2)/t.

Plugging in the values given, we get v0 = (30 m - 1/2*(9.8 m/s^2)*(2.5 s)^2)/(2.5 s) = 12 m/s. This is the magnitude of the initial velocity vector.

To find the angle above the horizontal, we can use the equation v0y = v0*sin(theta), where v0y is the vertical component of the initial velocity and theta is the angle above the horizontal. We can also use the equation v0x = v0*cos(theta), where v0x is the horizontal component of the initial velocity.

Since the ball is launched from ground level, we can assume that the initial vertical velocity is 0, so v0y = 0. This means that v0*sin(theta) = 0, which can only be true if either v0 = 0 or sin(theta) = 0. Since the initial velocity cannot be 0, we can conclude that sin(theta) = 0, which means that theta = 0 degrees or 180 degrees. This makes sense, as the ball could have been launched at a 0 degree angle (straight ahead) or a 180 degree angle (straight back).

To determine which angle is correct, we can use the equation v0x = v0*cos(theta) and plug in our calculated value for v0. We get v0x = 12 m/s*cos(theta). Since the ball lands 30 m away, we know that the horizontal component of the initial velocity must be 30 m (since there is no acceleration in the horizontal direction). This means that 12 m/s*cos(theta) = 30 m, which can only be true if cos(theta)